Functions, operator => eigenfunction, eigenvalue

sundriedtomato

1. The problem statement, all variables and given/known data
Show, that functions
f1 = A*sin([tex]\theta[/tex])exp[i[tex]\phi[/tex]] and
f2 = B(3cos[tex]^{2}[/tex]([tex]\theta[/tex]) - 1) A,B - constants
are eigenfunctions of an operator

and find eigenvalues


3. The attempt at a solution
This is what i got for the first function:


The next step is to solve left part of the equation, and than compare it to the right part.

The question arises, how to solve that equation?

I tried simplifying left part of an equation in mathcad, and I got


Next question from that part, is if I am doing it right, how to compare those parts, and answer a question - weather this function is an eigenfunction of an operator?

Thank You in advance, and I am constantly near computer and waiting for suggestions.

Reshma

Your eigen operator has partial derivatives wrt to [itex]\theta[/itex] and [itex]\phi[/itex]. When you operate it on your given eigen function, you should get back your original function multiplied by a scaling factor which is your eigenvalue.

sundriedtomato

Yes, Thank You, but how do I calculate that? May I ask for instructions on how to calculate that left part of an equation? Is the result I got is correct?

nrqed

Just go ahead and apply the derivatives!! It's that simple. (btw, I don't know what you entered in mathcad but what it gave you is wrong).

all you have to do is to apply the derivatives

nrqed

For the first term, what you have to calculate is

[tex] \frac{1}{sin \theta} \frac{\partial}{\partial \theta} ( sin \theta ~\frac{\partial}{ \partial \theta} (A sin \theta} e^{i \phi}}) ) [/tex]

sundriedtomato

I will give it a try right know. Thank You.

sundriedtomato

So, after proper calculations, the result is


Is this one correct?

I posted an image of what I am given, and as far as I know, differentiation sign usually is placed before the function?
I just don't get it - to what parts of an equation do underlined derivatives belong t?


Thank You.

rahuldandekar

The part in brackets is an "operator"... every incomplete differentiation sign (ie, without anything to differentiate) operates on whatever is "multiplied" to it.

For example, the d/d(theta) is your image also operates on A*sin(theta)*exp(i*phi), because when you open the bracket, it gets to differentiate that term.

Eg. (the d's are partial)
[d/dx + d/dy]*x*y^2 = y^2 + 2*y*x

sundriedtomato

....

Reshma

Just operate it on the function. They are simple partial derivatives, I won't take much time to solve. Just a little patience for the initial steps. I tried your problem and many terms get canceled out and you get the solution correctly. You don't need mathcad to do it.

sundriedtomato

Thank You Reshma! I already managed to solve it correctly, and find eigenvalues as well.
For the case with the first function : a = 2h^2, and for the case with second function b = 6h^2. The problem was initially in understanding how to apply operator properly. Once example have been given, and operator properties reanalyzed - problem got solved in like 10 minutes (with putting it all on a paper as well). Thank You everybody who took part in this one!